# When does a mathematical predicate have a truth value?

Say we have a predicate in a domain of real numbers, P(x), 2x+10=20 we know that we can existentially quantify this and say that the value x=5 makes this true, but we cannot talk about P(x) being true unless we know x, and hence we can't talk about it's truth value unless we talk about an assignment for x, (if we replace x with the name of an object) or we bind x using quantifiers to talk about what conditions it will be true under.

If we have a different situation where the predicate P is x+x=2x we know we can universally quantify this statement, would we say x is true without x necessarily having an assignment as this statement is always true, or does the statement only have a truth value when we talk about a value.

• A predicate P(x) is true for some "values" assigned to variable +x* and false for others. If the quantified formula is for all x Px, then it is true for some interpretation and false for others. May 4, 2022 at 5:58
• I cannot produce a statement like \$x+x=2x\$ and simply say that 'this is true' more that for any number \$x\$, this will be true. @MauroALLEGRANZA? May 4, 2022 at 11:02
• Not clear... Is x+x=2x a "predicate" (open formula)? Yes. Assign to x e.g. the value 3 and we get 3+3=2.3 which is true. Is the sentence ∀x (x+x=2x) in the domain of e.g. natural numbers? Yes it is. May 4, 2022 at 11:22
• no, I might be misunderstanding however. May 4, 2022 at 13:37
• I didn't explain myself that well actually, when we write P(3) as having a truth value, it is the same as x+3=6 when x is assigned 3, but what I'm asking is, if under no assignment the statement 'x+3=6' has no truth value we can talk about, without using a quantifier? May 4, 2022 at 13:41

This is an insight going back at least to Frege: a sentence/statement/proposition (not the same clearly, but the rough notion is there) is the bearer of truth. Hence predicates do not have truth value. This follows from the traditional model theoretic semantics- a predicates reference is a function from the domain into the truth values.

• Are you saying that predicates themselves don't have truth values? Or that their values for certain members have truth values like P(1) but P(x) where x is a free variable (or just p) does not? May 4, 2022 at 11:11
• "Predicates do not have truth value". P(1), P(2), etc... are not predicates. It is perhaps instructive to look at these terms in the lambda calculus. Using beta- reduction, they are equivalent value wise to truth. Meanwhile, the predicate \x. 2x +10 = 20 is already maximally reduced. May 4, 2022 at 16:48
• thank you very much, almost like 'the predicate has no truth value' but application of the predicate to a value does. May 5, 2022 at 8:56

1. Since both statements are statements of real numbers, the predication of truth is a function of axioms, theorems and proof of reals; establishing whether or not the veracity of a mathematical statement is false, conditionally true, or universally true is in essence one of the major functions of mathematical proof. Hence, when an algebraic statement is true, we know because we prove (direct proof, for example) or disprove (proof by contradiction, for example).

2. In the first case, we know the claim `2x+10=20` is conditionally true because we use a series of algebraic axioms to reduce the statement to `x=5` [particularly by inverting the operations and their order such that if our original statement is some quantity subjected to operations `(*2,+10,=20)`, then it is true in any cases residual of applying operations `(=20,-10,/2)`]. So, we do the algebra and see what we got. In fact, the two statements combined can be shown consistent because they instantiate the identity axiom `a=a` as `20=20`. Thus, the truth of the statement S1(x) is conditional on x.

3. We have exactly the same challenge in the second, and instead find by simplifying the RHS to `2x=2x`. No matter how hard we try to simplify the equation, we will not be able to isolate the variable. As such, any value works by direct proof, again invoking the identity axiom `a=a` which we presume to be true. (It would be messy otherwise.) Thus, the statement S2(x) is tautological.

4. Let's do a third example: `x + x + x = 2x`. Here, if we simplify and then divide both sides by quantity (barring 0 as a value), it reduces to 3=2. This violates the identity axiom `a=a`. So, since `3=2` contradicts `a=a` we must reject either the axiom or the conclusion, and by extension the original statement. Thus, while we can create the appearance of a predication of some quantity of the reals, using the rules that govern reals, we find the predication is always false. Thus, the predication S3(x) is contradictory and necessarily false (if we want to hold on to our axioms).

5. Lastly, the truth value of a predication may not be possible to be proven within the axiomatic formal system from which it arises. A famous example is the Goldbach Conjecture which states that:

every even whole number greater than 2 is the sum of two prime numbers. The conjecture has been shown to hold for all integers less than 4 × 10^18, but remains unproven despite considerable effort.

So, S4(n,p1,p2) which is `2n=p1+p2` for n in naturals, and p1 and p2 in primes seems to be true, but no proof exists for the predication, and no counterexample has arisen either. Thus, it is unknown whether this predication is conditional or tautological.

### Conclusion

Don't let the philosophical jargon predication complicate your understanding. In mathematics, predication is nothing more than binding a variable to a domain and then predicating the truth-condition on the consistency of the statements within an axiomatic system, and I suspect you've been doing it by at least intuition since elementary school. We know when mathematical predications are true when we devise mathematical proofs, disproofs (as in the irrationality of 2), or when we discover counterexamples. And ultimately to become a professional mathematician is to fully explore both proof theory and model theory.